Reading files from a folder using python

Question

I am trying to read all files within a directory.

for root, dirs,files in os.walk(path):
        for j in files:
            print(str(j))

This is my code where path is the path of the directory to be read.But it does not print the files in the order of their names. In my case, I have files from 0.txt,1.txt,.... to 3590.txt. I want it to print the files in the same order. But rather it starts from 579.txt . How can I fix this? I want to do some computation on the files in the same order and so just storing the names in a list and sorting it wont help.

"I want to do some computation on the files in the same order and so just storing the names in a list and sorting it wont help." You can use a list and still perform computations on its contents. Can you explain further why this wouldn't work? — Kevin
– Kevin, Commented Sep 28, 2015 at 12:34
Related to Nonalphanumeric list order from os.listdir() in Python — Rodrigue
– Rodrigue, Commented Sep 28, 2015 at 12:40

RickyA · Accepted Answer · 2015-09-28 12:46:31Z

4

what about sorting them with a lambda to use the int in the filename:

for root, dirs,files in os.walk(path):
    for j in sorted(files, key=lambda key: int(key.replace(".txt", ""))):
        print(str(j))

edited Sep 28, 2015 at 12:46

answered Sep 28, 2015 at 12:37

RickyA

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5 Comments

Kevin Over a year ago

This won't sort them numerically, though. For example, sorted(["2.txt", "11.txt", "300.txt"]) gives ['11.txt', '2.txt', '300.txt']. (this may actually be what the OP means by "sorting won't help")

Noober Over a year ago

No it is not working as Kevin said. I am getting something like this-539.txt 54.txt 540.txt 541.txt

Kevin Over a year ago

If lexicographic sorting was really the problem though, I wouldn't expect to see "539.txt" first, since "1.txt" and "0.txt" still precede it alphabetically.

Noober Over a year ago

@Kevin Yes 0.txt and 1.txt appear before 539 . I just posted three files where it is failing.In actual output,0 and 1 appear before 539

J. Corson Over a year ago

If you want the sorting to be in numerical order rather than lexicographic, you could use the natsort library. from natsort import natsorted then replace sorted in the above code with natsorted.

Vivek Sable · Accepted Answer · 2015-09-28 12:43:46Z

0

As file name structure is like $Number$.txt, so we can sort by Number.

Demo:

>>> l = ["0.txt", "1.txt", "33.txt", "2.txt", "10.txt", "11.txt"]
>>> sorted([(int(i.replace(".txt", "")), i) for i in l ])
[(0, '0.txt'), (1, '1.txt'), (2, '2.txt'), (10, '10.txt'), (11, '11.txt'), (33, '33.txt')]
>>> [i[1] for i in sorted([(int(i.replace(".txt", "")), i) for i in l ])]
['0.txt', '1.txt', '2.txt', '10.txt', '11.txt', '33.txt']

edited Sep 28, 2015 at 12:43

answered Sep 28, 2015 at 12:35

Vivek Sable

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Comments

Irfan __s · Accepted Answer · 2024-03-04 09:13:13Z

0

you can try sorting them by using natsort:

import natsort
unsorted_files = ['2.txt','3.txt','1.txt']
sorted_files = natsort.natsorted(unsorted_files)
print(sorted_files)

you need to install the package if not installed by using :

pip install natsort

edited Mar 4, 2024 at 9:13

answered Mar 4, 2024 at 9:09

Irfan __s

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